So confused... empirical and molecular formulas?

Question

compound contains only C, H, and N. Combustion of 42.0 mg of the compound produces 40.2 mg CO2 and 49.3 mg H2O. What is the empirical formula of the compound?

Answer 1

Moles CO2 = 0.0402 g/ 44.0098 g/mol =0.000913

Mass C = 12.011 g/mol x 0.000913 mol = 0.0110 g

Moles water = 0.0493 g / 18.02 g/mol = 0.00274

Moles H = 2 x 0.00274 = 0.00548

Mass H = 0.00548 mol x 1.008 g/mol =0.00552 g

Mass N = 0.042 - ( 0.0110 + 0.00552) = 0.0255 g

Moles N = 0.0255 / 14.0067 g/mol = 0.00182

C (0.000913) H ( 0.00548) N ( 0.00182)

we divide by the smallest number to get the empirical formula

0.000913/ 0.000913 = 1 => C

0.00548/ 0.000913 = 6 => H

0.00182 / 0.000913 = 2 => N

CH6N2